The main result of the paper is that the gravitational effects are proportional to the energy density of the electromagnetic field inside the cavity (roughly speaking). This means that the electromagnetic field inside the cavity determines the way the gravitational field is distributed. A change could be due to the proportionality factor L(x). Of course, there could be an effect due to the Earth gravitational field but I did not estimate it yet.

So, if I understand correctly, L(x) is purely a function of the position "x vector" obtained by a defined volume integral, and is only dependent on the geometry of the cavity and the l0 (ell zero) constant which depends on U0 which depends on the source energy.

In Equation 37, given a defined geometry of the frustum, "a" is later determined to be a constant term, "b" is also a constant term, so the shape of the function depends on the logarithm and the square of "r vector". However Equation 38 sets a condition by which "r vector" is linear to "z vector". Does this mean that "r vector" is a function of z, i.e. it varies with the position on the height axis of the frustum?Am I understanding it correctly?

If that is the case, then the gravitational effect is stronger towards one plate than the other. Which leads to my concerns about a torque being present rather than a thrust. Presumably, in an experiment the frustum is suspended or supported in such a way that a static balance is achieved; however if a torque is present the balance is altered and this could potentially be confused for a thrust depending on how the mechanical parts of the experiment are set up.(I've realised that there could be a torque even in vacuum because you can't remove the existence of the frustum itself, and its weight distribution could change with the gravitational effect)

But again, I'm not 100% sure I'm understanding this correctly. I'm just throwing ideas for the smarter people to think about. I hope this is useful.

Yes, I have to impose the condition that I am living in a frustum to solve my equation for L(x). So, you can express it both as function of r or z. Please, note that l0 (ell zero) is very large making some contributions less important than others. Besides, contributions from Earth gravitational field are damped by a factor (Schwarzschild radius)/(Earth radius) where the Schwarzschild radius of the Earth is 9 mm while the Earth radius is about 6000 km. This factor being 10^(-12) makes negligible the effect with respect to the behaviour of the electromagnetic field inside the cavity.

I remain unconvinced that calculating resonant modes for cylinders is a good approximation for conical frustums though.

The problem as I see it with the cylindrical resonate model to the (in progress) frustum resonate model is in the cylinder model the guide wavelength / frequency stays the same from one end to the other, while in the frustum, the guide wavelength / frequency continually varies as the diameter varies.

These are guide wavelength / frequency and NOT cut off wavelength/frequencies. At 2.45GHz, both of your ends operate well above cutoff.

So for your frustum, operating at 2.45GHZ, there is a 181.3MHz difference in the guide frequencies at each end.

Trying to imagine what this would look like. Can easily see a cylindrical model with constant guide wavelength from end to end.

I may be wrong but as I see it cavity resonance occurs at the internal guide wavelength and not at the external applied Rf wavelength. We know the guide frequency at each end of the frustum, so how to get the applied Rf frequency that causes end plate to end plate resonance at 1/2 wave or other harmonic of some internal guide frequency?

Besides, contributions from Earth gravitational field are damped by a factor (Schwarzschild radius)/(Earth radius) where the Schwarzschild radius of the Earth is 9 mm while the Earth radius is about 6000 km. This factor being 10^(-12) makes negligible the effect with respect to the behaviour of the electromagnetic field inside the cavity.

I see! So there may be an effect but the entity would be too small compared with the values in the experiments.

I remain unconvinced that calculating resonant modes for cylinders is a good approximation for conical frustums though.

The problem as I see it with the cylindrical resonate model to the (in progress) frustum resonate model is in the cylinder model the guide wavelength / frequency stays the same from one end to the other, while in the frustum, the guide wavelength / frequency continually varies as the diameter varies.

These are guide wavelength / frequency and NOT cut off wavelength/frequencies. At 2.45GHz, both of your ends operate well above cutoff.

So for your frustum, operating at 2.45GHZ, there is a 181.3MHz difference in the guide frequencies at each end.

Trying to imagine what this would look like. Can easily see a cylindrical model with constant guide wavelength from end to end.

I may be wrong but as I see it cavity resonance occurs at the internal guide wavelength and not at the external applied Rf wavelength. We know the guide frequency at each end of the frustum, so how to get the applied Rf frequency that causes end plate to end plate resonance at 1/2 wave or other harmonic of some internal guide frequency?

You can get a reasonably close approximation to the exact expression for the natural frequencies of a truncated cone

What is the "exact expression for the natural frequencies of a truncated cone"?

Why go with close estimates?

The exact expression for the solution to Maxwell's equations in a truncated cone involves the solution of two eigenvalue problems, one in terms of Legendre Associated functions and the other one in terms of spherical Bessel functions. Take a gander at this http://gregegan.customer.netspace.net.au/SCIENCE/Cavity/Cavity.html for an exact expression (Egan only gives the exact solution for TE0np and TM0np modes, not valid for TM212 for example), so that you can see that it would be unwieldly to program it in Excel.

Minotti also gives the exact solution to Maxwell's equations in a truncated cone (section 4 of http://arxiv.org/pdf/1302.5690v3 ) quoting Greg Egan. The exact solution goes back to the 1930's in papers by S.A. Schelkunoff of Bell Labs.

Here is a picture of Schelkunoff examining some conical cavities in the early 1930’s. (Bad Joke coming: They seem to be tied down, maybe that's why Bell Labs didn't report any thrust from conical cavities )(Image from P. C. Mahon, Mission Communications: The Story of Bell Laboratories, 1975.)

That's what I'm talking about! Something like this would be at least 20% more efficient than an EM Drive could ever be. IMO, no competition...

Todd D.

The concept of the EM Drive (if the EM Drive is ever possible) is self-contained while the Photonic Thruster is limited to the distance to the Resource Vehicle, at the present time:

Quote from: http://www.spaceref.com/news/viewpr.html?pid=45847

I can see future development that includes optical cavities that span many kilometers achieved with precise mirror alignment to enable maneuvering spacecraft many kilometers apart, and propellant-free propulsion of satellites in formations.

Although Bae has written about the concept being used for Interstellar Travel:

Does the Flight Thruster have a slightly concave top and convex bottom? Would appear so from the gaps.

Enhanced the photo as much as I can for those wishing to try to extract dimensions as this photo is better that the original as it has no distortion.

If we can find the dimensions of the bottom Rf connector flange, we can set pixels per cm and start doing measurements.

Most N connectors like that are 1" square, and the holes are .718" center line to center line.

Pixel away.

Thanks. Have fine rotated to vertical / horizontal and lined up. Attached if anyone else wants to have a go.

If you tell me specifically what measure you are trying to find, and what measures you know with regards to the photo, I might be able to help.

The size of the rectangular section the Flight Thruster is sitting on and the size of the Rf connector flange and mounting hole spacing should be knowable.

Assuming the build has the concave and convex spherical section machined in the top and bottom plates, the internal cavity height is then the distance from the underside lip of the top plate to the upper lip of the bottom plate.

Looking at the surface of the Flight Thruster, it looks either cast or machined but not rolled, so assume a 1mm thick wall. Should then be able to determine the internal diameter at the bottom and top. Like lifting off the end caps and measuring the inside top and bottom diameters.

Here is an enhanced, better resolution than the original, image of the Flight Thruster to work from. I found the image had to be rotated 1.5 deg CCW to bring the bottom alum beam into horizontal. I can do that for you if you need.

BTW thanks. Your efforts on working out the dimension from the 1st Flight Thruster photo were fantastic. This non distorted and higher resolution image, with more knowable dimensions, should be easier to work with.

Updated image with a better one.

Looking at the image again, there may be copper end plates that have the curve and those end plates are sandwiched between a upper and lower flange that is part of the Flight Thruster case and bolted on end caps that are also internally curved to allow the copper end plates curves to exits without being damaged.

I do however note the bottom end cap doesn't appear to be sitting flush on the bottom beam, maybe it has a convex curve on the bottom and likewise the top end cap appears to be concaved.

Have attached my effort at making up the Flight Thruster. Note it is rotated 1.5 deg CCW.

The size of the rectangular section the Flight Thruster is sitting on and the size of the Rf connector flange and mounting hole spacing should be knowable.

The one "known" is the 1" N connector. Notice that it is slightly off-centre because 1) the top is more distant than the bottom (frustum angle) and 2) it does not lie exactly face-on to the image plane, but rather a little around the circumference to the left. Fortunately we can correct for both of these small changes fairly exactly.1) The frustum angle is directly measurable2) Since both circumferential edges are visible, the degree of rotation is directly computable.Having converted the "one inch, modified" to pixels it's off to the races.

Yesterday night i did a new test with the Magnetron moved to the small side (10cm from the small side). I patched the previous hole.I also put a coil around one magnet in hope to change the frequency.I ordered a frequency counter, so i will now exactly what is the frequency produced and the intensity.

No pendulum movement was observed. The duration of the test was ~40 sec.

In the future tests i will be able to observe any change in frequency by modifying the current in the coilIf this will not change the frequency i will modify the frustum to add a moving plate inside.

Moving a small internal end plate back and forth was what Shawyer seemed to do to get his cavity into length resonance.

As per attached.

Would suspect the movable small end plate was very near or at the right end of the cylinder as in the drawing.

Yes this is the disk i`m refering. I read almos all this thread and first one, and almost all the papers related to em drive from www.emdrive.comI bought this counter,i should receive it in a few days.

Be cautious of these broadband Frequency counters. My pesonal experience is that they may register a spur frequency because it could be more sensitive in the UHF range as opposed to over 2 GHz. The only accurate way to analyze the Fc of the magnetron would be either a Spectrum Analyzer or filtered frequency counter. Magnetrons are loaded with many frequencies.

Okay, I think I can give you the numbers you need. But I'm not sure I understand you. Are you saying you want the diameters of the end plates and the height of the cavity? And if you want the height of the cavity, do you mean excluding the concavities in the endcaps since we don't know their depth?

Also, if the 1" N connector is there in front, where is the RF connector and mounting hole? Is that the same as the N connector? If so, where does it measure 1", specifically?

And are you saying you also want the width of the thin metal beam the entire thruster is sitting upon?

I don't mean to be obtuse, but I'm not an engineer and this stuff is largely foreign to me.